Two resistors connected to a battery

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Homework Help Overview

The discussion revolves around a circuit problem involving two resistors, a 100-ohm and a 200-ohm resistor, connected in parallel to a 10-volt battery. Participants are tasked with determining the power dissipated by the 100-ohm resistor and calculating the current in the 200-ohm resistor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of relevant equations for power and current in the context of resistors in parallel. There are attempts to clarify the correct resistor for the power calculation and discussions about using Ohm's Law for the second part of the problem.

Discussion Status

The discussion is active with participants providing guidance and corrections to each other. Some have offered suggestions on how to approach the calculations, while others are clarifying misunderstandings about the problem's requirements.

Contextual Notes

Participants have noted confusion regarding the application of formulas and the specific resistors involved in the calculations. There is also a mention of a follow-up question regarding the same resistors connected in series, indicating an exploration of different circuit configurations.

jay_buckets
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Homework Statement
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.

Determine the power dissipated by the 100.-ohm resistor.

Calculate the current in the 200.-ohm resistor.
Relevant Equations
P=VI
P=I^2R
P= V^2/R
I honestly have no idea how to do this problem whatsoever. My teacher didn't show an example like this before.
 
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jay_buckets said:
Homework Statement:: In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.

Determine the power dissipated by the 100.-ohm resistor.

Calculate the current in the 200.-ohm resistor.
Relevant Equations:: P=VI
P=I^2R
P= V^2/R

I honestly have no idea how to do this problem whatsoever. My teacher didn't show an example like this before.
Welcome to the PF.

When the two resistors are connected in parallel to the battery, you can treat them as independent. Does that help? You have written the correct equations -- try applying them to the first question and show your work please. Thank you.
 
P= (10v^2)/(200.ohm) = 0.5 W. is that the correct answer for part 1?
 
jay_buckets said:
P= (10v^2)/(200.ohm) = 0.5 W. is that the correct answer for part 1?
I think part 1 asks about the 100 Ohm resistor, no? And good start, BTW. :smile:
 
ohhhh yeah it does oops. So would it be 1 W then?
 
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Yep. How about part 2 now? :smile:
 
would you use 1W=(I^2)100 ohms and then 1W/100ohms= I^2?
 
wait no would you calculate the power dissipated by the 200 ohm resistor and then use that formula?
 
For part 2, I would just use the simple form of Ohm's Law, V-IR.
 
  • #10
oh yeahhh thank you for your help!
 
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  • #11
You're welcome. And for bonus points, can you answer the same two questions for the case where the two resistors are connected in series across the 10 Ohm Volt battery? Hint -- start by calculating the series current by using the total series resistance and the battery voltage... :smile:
 
Last edited:
  • #12
berkeman said:
across the 10 Ohm battery?
Ohms, or Volts, or both?
 
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  • #13
Oops, thanks. Fixed it now. o0)
 

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